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Name:_____________
Chemistry 460 Biochemistry
First Hour Exam
Remember- Show all work for partial credit
1. (10 points) Name as many organelles from a eukaryotic cell as you can and give a 1
sentence description of the organelles function.
1 point / organelle and 1 point / explanation
Nucleus - membrane enclosed region that isolates the genetic material from the rest of the
cell
Mitochondria - organelle used to extract energy from chemical decomposition
Chloroplast - organelle used to obtain energy and make C 6H12O6 and O2 from H2O and
CO2 using light energy
lysosome - organelle used to digest material endocytosed into the cell
rough ER - membrane sac with ribosomes attached used to synthesis of proteins to be
exported out of cell
smooth ER - membrane sac distal from rough Er, proteins inside Er have now been
synthesized and are now being processed, lipid synthesis.
Golgi apparatus - used in final processing of secretory proteins
Peroxisome - membrane bound organelle used to isolate chemical reaction involving
peroxide
Glyoxisome - plant equivalent of peroxisome
Several other organelles accepted
2. (10 points) Some proteins contain added chemical groups or cofactors that give them
added chemical reactivity. One such cofactor is NAD+ or nicotinamide adenine
dinucleotide. Our body cannot synthesize this cofactor so it must be obtained in our diet
as the vitamin riboflavin. Below is the structure of NAD+. In this structure identify as many
different chemical functional groups as you can:
2 points/group, 5 minimum
OH on sugar alcohol
C=ONH2 amide or amido
aromatic rings with nitrogen to make nitrogen bases
PO4H - phosphate groups
O
O between two PO4's - anhydride linkage
NH2 primary amine
C
+
O in sugar ring look like an ether
O CH2 O N
NH2
- H H
HO P O
HO
O
H
P
OH
O-
O
CH2
H
OH
N
O
N
H H
H
H
OH OH
NH2
N
N
2
3. (10 points) I have said in class several times that because of its side chain pKa histidine
can be used as an acid or a base at neutral pH. Assume you have a .05M solution of
histidine, what are the molar concentration of both the acid (HA) and the base (A -) forms of
this amino acid at pH 7.00. (Ignore the pKa’s of the main chain functional groups to simplify
this problem.)
Ignoring the main chain functional groups mean that you just have to worry about the
side chain imidazole ring with a pKa of 6.00
I’ll call the acid from HisH+ and the base form His
HisH+ + His = .05; His = .05 - HisH+
pH = pKa + log A -/HA = pKa + log Base/acid= pKa + log His/HisH+
7.00 = 6.00 + log His/HisH+
1.00 = log His/HisH+
101 = His/HisH+
10 = His/HisH+
10(HisH+) = His = .05 - HisH+
11HisH+ = .05
HisH+ = .05/11 = .00455 = HA = His H +
His = .05-.0045 = .04545 = A-
3
4. (20 points) Fill in the following table:
Name
Isoleucine
Tyrosine
Lysine
3 letter
abbreviation
Ile
Tyr
Lys
1 letter
abbreviation
I
Y
K
Structure
OH
CH3
NH3+
CH2
CH2
H
NH3+
CH3
CH2
C COO-
CH2
2
H
CH2
NH3+ 11C COO-
CH2
NH3+
H
C COO-
15
H
side chain
pKa
(if ionizable)
-
10.07
10.53
General
classification
(nonpolar,
polar, etc)
Nonpolar
aromatic, polar
+ Charged
Base
5. (10 points) Calculate the pI of the peptide nh2-A-V-D-E-W-cooh
pKa’a nh2 about 9
cooh about 3
COOH of D 3.65
COOH of E 4.25
So I will have a titration curve with 4 flat regions: 3.3.65,4.25 and 9
Note: I am using a high pKa of 3 for my cooh end because this is not an amino acid so the
Charge effect from the nh3 end is minimal.
At the lowest pH I will be fully protonated with 3 COOH and 1 NH3+ for a +1 charge
I will therefor have a 0 charge at the end of my first flat region and will be at my pI
the pI must then be at the average of my first two flat regions
pI = (3+3.65)/2 = 3.325
Note your answer may be different depending on what you chose for the pKa
of the cooh terminus
4
6A. (10 points)Suggest a chromatographic method to separate the above peptide
(AVDEW) from the peptide AVRKW. Justify your choice of method.
From problem 5 above we known that the pI of AVDEW is low, to at pH 7 the
peptide will have a negative charge. The big difference is that the peptide AVRKW has
two basic side chains instead of two acidic side chains, so I would guess that it is +
charged at pH 7 instead. Thus ions exchange is the best method for separating these
peptides. At ph 7 peptide AVDEW will stick to a + charged column (an anion excharger)
but the peptide AVRKW will pass right through. Conversely if we used a cation exchanger
(- charged column) AVRKW will stick, and AVDEW will pass through
There is not a big enough difference in molecular weight for a gel column to work.
6B. (5 points) How could you tell when the above peptides eluted off the column?
How could you tell? Both peptides have a tryptophan. The tryptophan gives them a
UV absorbance, so all you have to do is to look at the absorbance of the solution around
280 nm.
6C. (10 points)Suggest a second chromatographic method you could use to separate the
above peptides from the protein cytochomre c (104 amino acids), and justify your answer.
Here we have a big difference in molecular weight so a gel permeation column
work be ideal.
Knowing that cytochrome c is an enzyme you might also suggest an affinity column
based on a compound that is known to bind to cytochrome c.
7. (5 points) Define primary, secondary, tertiary and quaternary structure in proteins.
Primary structure: The covalent structure of a protein , ie the sequence of amino acids and
the location of any disulfides.
Secondary structure: local structures that occur between a few residues
Tertiary structure: the overall structure of a single protein
Quaternary structure: The structure of porinte complexes containing more than one protein.
5
8.(10 points) Lets say that a probe has just come back from Mars, and this probe captured
a Martian life form that contained the following amino acid. Discuss the similarities of
differences between this Martian amino acid and its nearest earth equivalent.
H
NH3+ C
+
CH2 NH3
COOI was trying to make this a D amino acid instead of an L amino acid. However, when I
double checked my key, I discovered that this is also an L. Since this amino acid has a
side chain with a primary amine I would call its nearest earth equivalent lysine. The biggest
difference between this and lysine is that there are fewer C between the main chain and
the charged base group so the pK may shift a little bit, and the NH3+ group much closer to
the main chain so it can’t move around as much.